Solution

$A= \lbrace -1,2,3 \rbrace $ and $B= \lbrace 1,3 \rbrace $

(i) $A \times B= \lbrace (-1,1),(-1,3),(2,1),(2,3),(3,1),(3,3) \rbrace $

(ii) $B \times A= \lbrace (1,-1),(1,2),(1,3),(3,-1),(3,2),(3,3) \rbrace $

(iii) $B \times B= \lbrace (1,1),(1,3),(3,1),(3,3) \rbrace $

(iv) $A \times A= \lbrace (-1,-1),(-1,2),(-1,3),(2,-1),(2,2),(2,3),(3,-1),(3,2),(3,3) \rbrace $

Solution

We have, $ \ P = \lbrace x: x<3, x \in N \rbrace = \lbrace 1,2 \rbrace \ $ And $ \ Q = \lbrace x: x \leq 2, x \in W \rbrace = \lbrace 0,1,2 \rbrace $

Now, $ \ P \cup Q = \lbrace 0,1,2 \rbrace \text { and } P \cap Q= \lbrace 1,2 \rbrace $

$(P \cup Q) \times(P \cap Q) = \lbrace 0,1,2 \rbrace \times \lbrace 1,2 \rbrace $

$\hspace{3cm}= \lbrace (0,1),(0,2),(1,1),(1,2),(2,1),(2,2) \rbrace $

$ \therefore \quad P \cup Q= \lbrace 0,1,2 \rbrace \ \text { and } \ P \cap Q= \lbrace 1,2 \rbrace $

Solution

We have,

$A = \lbrace x: x \in W, x<2 \rbrace = \lbrace 0,1 \rbrace $,

$B = \lbrace x: x \in N, 1<x<5 \rbrace $ $ = \lbrace 2,3,4 \rbrace $

And $C= \lbrace 3,5 \rbrace $

(i) $A \times(B \cap C)$

$B \cap C= \lbrace 3 \rbrace $

$\therefore \quad A \times(B \cap C)= \lbrace 0,1 \rbrace \times \lbrace 3 \rbrace = \lbrace (0,3),(1,3) \rbrace $

(ii) $\quad A \times(B \cup C)$

$\because\quad(B \cup C)= \lbrace 2,3,4,5 \rbrace $

$ \begin{aligned} \therefore \quad A \times(B \cup C) & = \lbrace 0,1 \rbrace \times \lbrace 2,3,4,5 \rbrace \\ \\ & = \lbrace (0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5) \rbrace \end{aligned} $

Solution

(i) We have, $(2a+b, a-b)=(8,3)$

$ \Rightarrow \quad 2 a+b=8 \text { and } a-b=3 $

[since, two ordered pairs are equal, if their corresponding first and second elements are equal]

On substituting, $b=a-3$ in $2 a+b=8$, we get

$\quad \ \ 2a+a-3 =8 \\ \\ \Rightarrow 3 a-3=8 \\ \\
\Rightarrow 3a =11 \\ \\ \Rightarrow a=\dfrac{11}{3} $

$ \text {Again, substituting a} = \dfrac{11}{3} \ \text{in} \ b=a-3, \ \text{we get}$

$b=\dfrac{11}{3}-3=\dfrac{11-9}{3}=\dfrac{2}{3} $

$\therefore \ a=\dfrac{11}{3} \ \text{and b}=\dfrac{2}{3}$

(ii) We have, $\quad \begin{pmatrix} \dfrac{a}{4}, a-2b\end{pmatrix}=(0,6+b) $

$\Rightarrow \dfrac{a}{4} =0 \Rightarrow a=0 $

And $\quad a-2 b =6+b $

$\Rightarrow \quad 0-2 b =6+b $

$\Rightarrow \quad -3 b =6 $

$\therefore \quad b =-2 $

$\therefore \quad a =0, b=-2$

Solution

We have, $A= \lbrace 1,2,3,4,5 \rbrace \ $ and $ \ S= \lbrace (x, y): x \in A, y \in A \rbrace $

(i) The set of ordered pairs satisfying $x+y=5$ is

$ \lbrace (1,4),(2,3),(3,2),(4,1) \rbrace $

(ii) The set of ordered pairs satisfying $x+y<5$ is

$ \lbrace (1,1),(1,2),(1,3),(2,1),(2,2),(3,1) \rbrace $

(iii) The set of ordered pairs satisfying $x+y>8$ is

$ \lbrace (4,5),(5,4),(5,5) \rbrace $

Thinking Process

First, write the relation in Roaster form, then find the domain and range of $R$.

Solution

We have,

$ \begin{aligned} R & = \lbrace (x, y): x, y \in W, x^{2}+y^{2}=25 \rbrace \\ \\ & = \lbrace (0,5),(3,4),(4,3),(5,0) \rbrace \end{aligned} $

Domain of $R=$ Set of first element of ordered pairs in $R$ $= \lbrace 0,3,4,5 \rbrace $

Range of $R=$ Set of second element of ordered pairs in $R$ $= \lbrace 5,4,3,0 \rbrace $

Solution

We have,

$R_1 = \lbrace (x, y) \mid y=2 x+7, \text { where } x \in R \text { and }-5 \leq x \leq 5 \rbrace $

$\text { Domain of } R_1 = \lbrace -5 \leq x \leq 5, x \in R \rbrace =[-5,5] $

For range of $R_1 , \ \ y =2 x+7 $

$\text { When } x=-5 \text {, then } $ $y =2(-5)+7=-3 $

$ \text { When } x=5 \text {, then } $ $y =2(5)+7=17 $

$\therefore \ $ Range of $R_1 = \lbrace -3 \leq y \leq 17, y \in R \rbrace =[-3,17]$

Solution

We have, $R_2= \lbrace (x, y) \mid x$ and $y$ are integers and $x^{2}+y^{2}=64 \rbrace $

Since, 64 is the sum of squares of 0 and $\pm 8$.

When $x=0$, then $y^{2}=64 \Rightarrow y= \pm 8$

$x=8$, then $y^{2}=64-8^{2} \Rightarrow 64-64=0$

$x=-8$, then $y^{2}=64-(-8)^{2}=64-64=0$

$\therefore \quad R_2= \lbrace (0,8),(0,-8),(8,0),(-8,0) \rbrace $

Solution

We have

$R_3= \lbrace (x,|x|) \mid x \text { is real number } \rbrace $

Clearly, domain of $R_3=R$

Since, image of any real number under $R_3$ is positive real number or zero.

$ \therefore \quad \text { Range of } R_3=R^{+} \cup \lbrace 0 \rbrace \text { or }[0, \infty) $

Solution

(i) We have, $h= \lbrace (4,6),(3,9),(-11,6),(3,11) \rbrace $

Since, $3$ has two images $9$ and $11$ So, it is not a function.

(ii) We have, $f= \lbrace (x, x) \mid x$ is a real number $ \rbrace $

We observe that, every element in the domain has unique image. So, it is a function.

(iii) We have, $g=\bigg \lbrace \left(n, \dfrac{1}{n} \right)\mid n $ $\text{is a positive integer }\bigg \rbrace $

For every $n$, it is a positive integer and $\dfrac{1}{n}$ is unique and distinct. Therefore, every element in the domain has unique image. So, it is a function.

(iv) We have, $s= \lbrace (n, n^{2}) \mid n$ is a positive integer $ \rbrace $

Since, the square of any positive integer is unique. So, every element in the domain has unique image. Hence, it is a function.

(v) We have, $t= \lbrace (x, 3) \mid x$ is a real number $ \rbrace $

Since, every element in the domain has the image 3 . So, it is a constant function.

Solution

Given, $f$ and $g$ are real functions defined by $f(x)=x^{2}+7$ and $g(x)=3 x+5$.

(i) $f(3)=(3)^{2}+7=9+7=16 \ $ And $ \ g(-5)=3(-5)+5=-15+5=-10$

$\therefore\quad f(3)+g(-5)=16-10=6$

(ii) $f \left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^{2}+7=\dfrac{1}{4}+7=\dfrac{29}{4} \ $ And $ \ g(14)=3(14)+5=42+5=47$

$\therefore \quad f \left(\dfrac{1}{2}\right)\times g(14)=\dfrac{29}{4} \times 47=\dfrac{1363}{4}$

(iii) $f(-2)=(-2)^{2}+7=4+7=11 \ $ And $ \ g(-1)=3(-1)+5=-3+5=2$

$\therefore \quad f(-2)+g(-1)=11+2=13$

(iv) $f(t)=t^{2}+7 \ $ And $ \ f(-2)=(-2)^{2}+7=4+7=11$

$ \therefore \quad f(t)-f(-2)=t^{2}+7-11=t^{2}-4 $

(v) $f(t)=t^{2}+7 \ $ And $ \ f(5)=5^{2}+7=25+7=32$

$ \therefore \quad \dfrac{f(t)-f(5)}{t-5}, \text { if } t \neq 5 $

$ \qquad \ =\dfrac{t^{2}+7-32}{t-5} $

$ \qquad \ =\dfrac{t^{2}-25}{t-5}=\dfrac{(t-5)(t+5)}{(t-5)} $

$\qquad \ =t+5 $

Solution

We have,

$ f(x)=2 x+1 \text { and } g(x)=4 x-7 $

$\text { (i) } \because \quad f(x)=g(x) $

$\Rightarrow \quad 2 x+1=4 x-7 $

$\Rightarrow \quad 2 x=8 $

$\therefore \quad x=4 $

$\text { (ii) } \because \quad f(x)<g(x) $

$\Rightarrow \quad 2 x+1<4 x-7 $

$\Rightarrow \quad 1+7 < 4x-2x$

$\Rightarrow \quad 8 < 2x$

$\therefore \quad x>4$

Solution

We have, $f(x)=2 x+1$ and $g(x)=x^{2}+1$

(i) $(f+g)(x)=f(x)+g(x)=2 x+1+x^{2}+1=x^{2}+2 x+2$

(ii) $(f-g)(x)=f(x)-g(x)=(2 x+1)-(x^{2}+1)=2 x+1-x^{2}-1=2 x-x^{2}=x(2-x) $

(iii) $(f g)(x)=f(x) \cdot g(x)=(2 x+1)(x^{2}+1)=2 x^{3}+2 x+x^{2}+1=2 x^{3}+x^{2}+2 x+1 $

(iv) $ \ \dfrac{f}{g}(x)=\dfrac{f(x)}{g(x)}=\dfrac{2 x+1}{x^{2}+1}$

Solution

We have,

$ f: X \rightarrow R, f(x)=x^{3}+1 $

Where

$X= \lbrace -1,0,3,9,7 \rbrace $

When

$x=-1$, then $f(-1)=(-1)^{3}+1=-1+1=0$

$x=0$, then $f(0)=(0)^{3}+1=0+1=1$

$x=3$, then $f(3)=(3)^{3}+1=27+1=28$

$x=9$, then $f(9)=(9)^{3}+1=729+1=730$

$x=7$, then $f(7)=(7)^{3}+1=343+1=344$

$f= \lbrace (-1,0),(0,1),(3,28),(9,730),(7,344) \rbrace $

$\therefore \quad$ Range of $f= \lbrace 0,1,28,730,344 \rbrace $

Solution

$ f(x)=g(x) $

$ \Rightarrow \quad 3 x^{2}-1 =3+x $

$ \Rightarrow \quad 3 x^{2}-x-4 =0 $

$ \Rightarrow \quad 3 x^{2}-4 x+3 x-4 =0 $

$ \Rightarrow \quad x(3 x-4)+1(3 x-4) =0 $

$ \Rightarrow \quad (3 x-4)(x+1) =0 $

$ \therefore \quad x =-1, \dfrac{4}{3}$

Thinking Process

First, find the two equation by substitutions different values of $x$ and $g(x).$

Solution

We have,

$ g= \lbrace (1,1),(2,3),(3,5),(4,7) \rbrace $

Since, every element has unique image under $g$. So, $g$ is a function.

Now, $ g(x) =\alpha x+\beta $

When $x=1$, then

$ g(1) =\alpha(1)+\beta $

$ 1 =\alpha+\beta \quad \ldots (i)$

When $x=2$, then

$ g(2) =\alpha(2)+\beta$

$ \Rightarrow \quad 3=2 \alpha+\beta \quad \ldots (ii) $

On solving Eqs. (i) and (ii), we get

$ \alpha=2, \beta=-1 $

Solution

(i) We have, $f(x)=\dfrac{1}{\sqrt{1-\cos x}}$

$\Rightarrow \quad -1 \leq \cos x \leq 1$

$\Rightarrow \quad -1 \leq -\cos x \leq 1$

$\Rightarrow \quad -1+1 \leq 1- \cos x \leq 1$

$\Rightarrow \quad 0 \leq 1- \cos x \leq 1$

Now, $f(x)$ is defined iff

$1-\cos x \neq 0$

$\Rightarrow \quad \cos x \neq 1$

$\Rightarrow \quad x \neq 2n \pi \quad \forall n \in Z$

$\therefore $ Domain of $f= R- \lbrace 2n \pi : n \in Z \rbrace $

(ii) We have, $f(x) =\dfrac{1}{\sqrt{x+|x|}} $

$x+|x| =$ $\begin{cases} x-x=0, & x<0 \\ \\ x+x=2 x, & x \geq 0 \end{cases}$

$ \because \quad x+|x|=x-x=0, x<0 $

Hence, $f(x)$ is defined, if $x>0$.

$ \therefore \quad \text { Domain of } f=R^{+} $

(iii) We have, $f(x)=x|x|$

Clearly, $f(x)$ is defined for any $x \in R$.

$\therefore$ Domain of $f=R$

(iv) We have, $ f(x)=\dfrac{x^{3}-x+3}{x^{2}-1} $

$f(x)$ is not defined, if

$x^{2}-1 =0 $

$(x-1)(x+1) =0 $

$x =-1,1 $

Domain of $f =R- \lbrace -1,1 \rbrace $

(v) We have, $ f(x)=\dfrac{3 x}{28-x} $

Clearly, $f(x)$ is defined, if

$28-x \neq 0$

$\Rightarrow \quad x \neq 28$

$\therefore \quad$ Domain of $f=R- \lbrace 28 \rbrace $

Thinking Process

First, find the value of $x$ in terms of $y$, where $y=f(x)$. Then, find the values of $y$ for which $x$ attain real values.

Solution

(i) We have, $f(x) =\dfrac{3}{2-x^{2}} $

Let $y =f(x) $

Then, $y =\dfrac{3}{2-x^{2}} $

$\Rightarrow 2-x^{2}=\dfrac{3}{y} $

$\Rightarrow \quad x^{2} =2-\dfrac{3}{y} $

$\Rightarrow \quad x=\sqrt{\dfrac{2 y-3}{y}}$

$x$ assums real values, if $2 y-3 \geq 0$ and $y>0 \Rightarrow y \geq \dfrac{3}{2}$

$ \therefore \quad \text { Range of } f=[\dfrac{3}{2}, \infty) $

(ii) We have, $f(x)=1-|x-2|$

We know that, $ \mid x-2 \mid \geq 0 $

$\Rightarrow \ \ -\mid x-2 \mid \leq 0$

$\Rightarrow \ \ $ $1- \mid x-2 \mid \leq 1 $

$\Rightarrow \ \ $ $f(x) \leq 1 $

$\therefore \ \ $ Range of $f=(-\infty , 1] $

(iii) We have, $f(x)=|x-3|$

We know that, $ |x-3| \geq 0 $

$ \Rightarrow \ \ f(x) \geq 0 $

$ \therefore \text { Range of } f=[0, \infty) $

(iv) We have, $f(x)=1+3 \cos 2 x$

We know that, $-1 \leq \cos 2 x \leq 1 $

$\Rightarrow \quad -3 \leq 3 \cos 2 x \leq 3$

$\Rightarrow \quad 1-3 \leq 1 + 3 \cos 2 x \leq 1+3 \\ \\ \Rightarrow \quad -2 \leq 1+3 \cos 2 x \leq 1+3 \\ \\ \Rightarrow\quad \ -2 \leq f(x) \leq 4 $

$\therefore \text { Range of } f =[-2,4] $

Thinking Process

First find the interval in which $|x-2|$ and $|2+x|$ is defined, then find the value of $f(x)$ in that interval.

Solution

Since, $\mid x-2 \mid = $ $\begin{cases} -(x-2), & x < 2 \\ \\ x-2, & x \geq 2 \end{cases}$

And $\mid 2+x \mid =$ $\begin{cases} -(2+x), & x < -2 \\ \\ 2+x, & x \geq -2 \end{cases} $

Given that

$f(x)= \mid x-2 \mid + \mid 2+x \mid , -3 \leq x \leq 3$

$f(x)=$ $\begin{cases} -(x-2)-(2+x), & -3 \leq x < -2 \\ \\ -(x-2)+(2+x), & -2 \leq x <2 \\ \\ (x-2)+(2+x), & 2 \leq x \leq 3 \end{cases}$

Or, $f(x)=$ $\begin{cases} -x+2-2-x, & -3 \leq x <-2 \\ \\ -x+2+2+x, & -2 \leq x < 2 \\ \\ x-2+2+x, & 2 \leq x \leq 3 \end{cases}$

Or, $f(x)=$ $\begin{cases} -2x, & -3 \leq x < -2 \\ \\ 4, & -2 \leq x <2 \\ \\ 2x, & 2 \leq x \leq 3 \end{cases}$

Solution

We have, $\quad f(x)=\dfrac{x-1}{x+1}$

(i) $f \left(\dfrac{1}{x}\right)=\dfrac{\dfrac{1}{x}-1}{\dfrac{1}{x}+1}=\dfrac{(1-x) / x}{(1+x) / x}=\dfrac{1-x}{1+x}=\dfrac{-(x-1)}{x+1}=-f(x)$

(ii) $f\left(-\dfrac{1}{x}\right)=\dfrac{-\dfrac{1}{x}-1}{-\dfrac{1}{x}+1}=\dfrac{(-1-x) / x}{(-1+x) / x}$

$ \Rightarrow f\left(-\dfrac{1}{x}\right)=\dfrac{-(x+1)}{x-1}$

Now, $\quad \dfrac{-1}{f(x)}=\dfrac{-1}{\dfrac{x-1}{x+1}}=\dfrac{-(x+1)}{x-1}$

$\therefore \quad f\left(-\dfrac{1}{x}\right)=-\dfrac{1}{f(x)}$

Solution

We have, $f(x)=\sqrt{x}$ and $g(x)=x$ be two function defined in the domain $R^{+} \cup \lbrace 0 \rbrace $.

(i) $(f+g)(x)=f(x)+g(x)=\sqrt{x}+x$

(ii) $(f-g)(x)=f(x)-g(x)=\sqrt{x}-x$

(iii) $(f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{({3}/{2})}$

(iv) $\left(\dfrac{f}{g}\right)x=\dfrac{f(x)}{g(x)}=\dfrac{\sqrt{x}}{x}=\dfrac{1}{\sqrt{x}}$

Solution

We have, $\quad f(x)=\dfrac{1}{\sqrt{x-5}}$

$f(x)$ is defined, if $x-5>0 \Rightarrow x>5$

$\therefore \quad$ Domain of $f=(5, \infty)$

Let $f(x)=y$

$\therefore \quad y=\dfrac{1}{\sqrt{x-5}} \Rightarrow \sqrt{x-5}=\dfrac{1}{y}$

$\Rightarrow \quad x-5=\dfrac{1}{y^{2}}$

$\therefore \quad x=\dfrac{1}{y^{2}}+5 $

$x \in (5, \infty )$ and $\dfrac{1}{y^2} > 0$

$\Rightarrow \quad y \in R^{+}$

Hence, range of $f=R^{+} $

Solution

We have, $ f(x)=y=\dfrac{a x-b}{c x-a} $

$ \begin{aligned} \therefore \quad f(y) & =\dfrac{a y-b}{c y-a}=\dfrac{a \left(\dfrac{a x-b}{c x-a}\right)-b}{c \left(\dfrac{a x-b}{c x-a}\right)-a} \\ \\ & =\dfrac{a(a x-b)-b(c x-a)}{c(a x-b)-a(c x-a)}=\dfrac{a^{2} x-a b-b c x+a b}{a c x-b c-a c x+a^{2}} \\ \\ & =\dfrac{a^{2} x-b c x}{a^{2}-b c}=\dfrac{x(a^{2}-b c)}{(a^{2}-b c)}=x \\ \\ \therefore \quad f(y) & =x \end{aligned} $

Hence proved.

Thinking Process

First find the number of element in $A \times B$ and then find the number of non-empty relation by using $2^{n(A \times B)}-1$

Solution

Option (d): We have, $n(A) =m \text { and } n(B)=n $

Now, $n(A \times B) =n(A) \cdot n(B) $

$\qquad n(A \times B) =m n$

$\therefore \quad $ Total number of relation from $A$ to $B=2^{n(A \times B)}-1=2^{m n}-1$

• Option (a) $ m^n $: This option represents the number of functions from set $ B $ to set $ A $, not the number of non-empty relations. A function is a specific type of relation where each element in $ A $ is related to exactly one element in $ B $. Therefore, this does not account for all possible non-empty relations.

• Option (b) $ n^m - 1 $: This option is incorrect because it represents the number of functions from $ A $ to $ B $ minus one. Similar to option (a), it only considers functions, not all possible relations.

• Option (c) $ mn - 1 $: This option is incorrect because it represents the number of non-empty pairs in the cartesian product $ A \times B $ minus one.

Thinking Process

If $a$ and $b$ are two successive positive integer and $[x]=a, b$, then $x \in [a, a+1 )$ & $x \in [b, b+1)$.

Solution

Option (d): We have, $ [x]^{2}-5[x]+6 =0 $

Let, $\quad[x] = z$

$\Rightarrow \quad z^2-5z +6 =0$

$\Rightarrow \quad z^2-3z-2z+6=0$

$\Rightarrow \quad z(z-3)-2(z-3)=0$

$\Rightarrow \quad z=2,3$

$\therefore \quad [x]=2$ and $[x]=3$

if $[x]=2$ then $x \in [2,3)$

if $[x]=3$ then $x \in [3,4)$

Combining the intervals, we get

$x \in [2,3) \cup [3,4)$

$\Rightarrow \quad x \in [2,4)$

• Option (a) $x \in[3,4]$: This option is incorrect because this intervals suggests that $x$ can take any value from $3$ to $4$ inclusive of both endpoints. $x$ cannot be in the interval $[3,4]$ as it would imply $[x] = 4$ which does not satisfy the equation.

• Option (b) $x \in(2,3]$: This option is incorrect because if $x \in (2,3]$, then $\left[x\right]$ would be 2 or 3. While $[x] = 2$ and $[x] = 3$ are solutions to the equation, the interval $(2,3]$ excludes $x = 2$, which is a valid solution. Therefore, the correct interval should include 2, making this option incorrect.

• Option (c) $x \in[2,3]$: This option is incorrect because if $x \in [2,3]$, then $\left[x\right]$ would be 2 or 3. For $[x]=3, x \in [3, 3+1)$. Therefore, this option is incorrect.

Solution

Option (c): We have, $\quad f(x)=\dfrac{1}{1-2 \cos x}$

We know that, $\quad -1 \leq \cos x \leq 1$

$\Rightarrow \quad -1 \leq - \cos x \leq 1$

$\Rightarrow \quad -2 \leq -2 \cos x \leq 2$

$\Rightarrow \quad 1-2 \leq 1-2 \cos x \leq 1+2$

$\Rightarrow \quad -1 \leq 1-2 \cos x \leq 3$

$\because \quad \cos x \neq \dfrac{1}{2}$ when $1-2 \cos x \neq 0$

$\Rightarrow \quad -1 \leq 1- 2 \cos x <0 \ $ & $ \ 0 < 1-2 \cos x \leq 3$

$\Rightarrow \quad -\infty < \dfrac{1}{1- 2 \cos x } \leq -1 \ $ & $ \ \dfrac{1}{3} \leq \dfrac{1}{1-2 \cos x} < \infty$

$\Rightarrow \quad - \infty < f(x) \leq 1 \ $ & $ \ \dfrac{1}{3} \leq f(x) < \infty$

$\therefore \quad$ Range of $f= \left(-\infty , -1\right] \cup \left[\dfrac{1}{3}, \infty\right)$

• Option (a) $\left[\dfrac{1}{3}, 1\right]$: $\left[\dfrac{1}{3}, 1\right]$ is incorrect because the range of $f(x)$ does not include 1. The correct range is $(-\infty , -1] \cup [\dfrac{1}{3}, \infty)$, and $1$ is not within this interval.

• Option (b) $\left[-1, \dfrac{1}{3}\right]$: $\left[-1, \dfrac{1}{3}\right]$ is incorrect because $f(x)$ can take values much larger than $\dfrac{1}{3}$ and less than $-1$.

• Option (d) $\left[-\dfrac{1}{3}, 1\right]$: $\left[-\dfrac{1}{3}, 1\right]$ is incorrect because the range of $f(x)$ does not include $-\dfrac{1}{3}$. The correct range is $(-\infty , -1] \cup [\dfrac{1}{3}, \infty)$, and $-\dfrac{1}{3}$ is not within this interval.

Solution

Option (c): We have,

$ \begin{aligned} f(x) & =\sqrt{1+x^{2}} \\ \\ f(x y) & =\sqrt{1+x^{2} y^{2}} \\ \\ f(x) \cdot f(y) & =\sqrt{1+x^{2}} \cdot \sqrt{1+y^{2}} \\ \\ & =\sqrt{(1+x^{2})(1+y^{2})} \\ \\ & =\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}} \end{aligned} $

$ \begin{matrix} \because & \sqrt{1+x^{2} y^{2}} \leq \sqrt{1+x^{2}+y^{2}+x^{2} y^{2}} \\ \\ \Rightarrow & f(x y) \leq f(x) \cdot f(y) \end{matrix} $

• Option (a) $f(x y)=f(x) \cdot f(y)$: This is incorrect because $ f(x y) = \sqrt{1 + x^2 y^2} \quad \text{and} \quad f(x) \cdot f(y) = \sqrt{(1 + x^2)(1 + y^2)} = \sqrt{1 + x^2 + y^2 + x^2 y^2} $

Since $\sqrt{1 + x^2 y^2} \neq \sqrt{1 + x^2 + y^2 + x^2 y^2}$, the equality does not hold.

• Option (b) $f(x y) \geq f(x) \cdot f(y)$: This is incorrect because $ f(x y) = \sqrt{1 + x^2 y^2} \quad \text{and} \quad f(x) \cdot f(y) = \sqrt{(1 + x^2)(1 + y^2)} = \sqrt{1 + x^2 + y^2 + x^2 y^2} $

Since $\sqrt{1 + x^2 y^2} \leq \sqrt{1 + x^2 + y^2 + x^2 y^2}$, the inequality $f(x y) \geq f(x) \cdot f(y)$ does not hold.

Solution

Option (b): Let $ f(x)=\sqrt{a^{2}-x^{2}} $

$f(x)$ is defined, if

$ a^{2}-x^{2} \geq 0 $

$\Rightarrow \quad x^{2}-a^{2} \leq 0 $

$\Rightarrow \quad (x-a)(x+a) \leq 0$

$\Rightarrow \quad -a \leq x \leq a$

$\therefore \quad \text {Domain of }f=[-a, a]$

• Option (a): $(-a, a)$ is incorrect because it excludes the endpoints $-a$ and $a$. The function $\sqrt{a^2 - x^2}$ is defined at $x = -a$ and $x = a$ since $a^2 - (-a)^2 = 0$ and $a^2 - a^2 = 0$, making the square root of zero valid.

• Option (c): $\left[0, a\right]$ is incorrect because it excludes the negative values within the interval $[-a, 0)$. The function $\sqrt{a^2 - x^2}$ is defined for all $x$ in the interval $\left[-a, a\right]$, not just the non-negative part.

• Option (d): $(-a, 0]$ is incorrect because it excludes the positive values within the interval $(0, a]$. The function $\sqrt{a^2 - x^2}$ is defined for all $x$ in the interval $\left[-a, a\right]$, not just the non-positive part.

Solution

Option (b): We have,

$ f(x) =a x+b $

$ f(-1) =a(-1)+b $

$ -5 =-a+b \quad \ldots (i)$

$ \text{ and,} \ f(3) =a(3)+b $

$ 3 =3 a+b \quad \ldots (ii)$

On solving Eqs. (i) and (ii), we get

$ a=2 \text { and } b=-3 $

• Option (a) $a=-3, b=-1$:

If $a = -3$ and $b = -1$, then:

For $f(-1) = -5$: $-3(-1) + (-1) = 3 - 1 = 2 \neq -5$

For $f(3) = 3$: $-3(3) + (-1) = -9 - 1 = -10 \neq 3$

Therefore, this option does not satisfy the given conditions.

• Option (c) $a=0, b=2$:

If $a = 0$ and $b = 2$, then:

For $f(-1) = -5$: $0(-1) + 2 = 2 \neq -5$

For $f(3) = 3$: $0(3) + 2 = 2 \neq 3$

Therefore, this option does not satisfy the given conditions.

• Option (d) $a=2, b=3$:

If $a = 2$ and $b = 3$, then:

For $f(-1) = -5$: $2(-1) + 3 = -2 + 3 = 1 \neq -5$

For $f(3) = 3$: $2(3) + 3 = 6 + 3 = 9 \neq 3$

Therefore, this option does not satisfy the given conditions.

Solution

Option (a): We have,

$ f(x)=\sqrt{4-x}+\dfrac{1}{\sqrt{x^{2}-1}} $

$f(x)$ is defined, if

$ \begin{aligned} 4-x & \geq 0 \text { or } x^{2}-1>0 \\ \\ x-4 & \leq 0 \text { or }(x+1)(x-1)>0 \\ \\ x & \leq 4 \text { or } x<-1 \text { and } x>1 \end{aligned} $

$ \therefore \quad \text { Domain of } f=(-\infty,-1) \cup(1,4] $

• Option (b): $(-\infty,-1] \cup(1,4]$ is incorrect because $x = -1$ is not in the domain of $f(x)$. At $x = -1$, the term $\dfrac{1}{\sqrt{x^2 - 1}}$ becomes undefined as the denominator becomes zero.

• Option (c): $(-\infty,-1) \cup[1,4]$ is incorrect because $x = 1$ is not in the domain of $f(x)$. At $x = 1$, the term $\dfrac{1}{\sqrt{x^2 - 1}}$ becomes undefined as the denominator becomes zero.

• Option (d): $(-\infty,-1) \cup[1,4)$ is incorrect because $x = 4$ is in the domain of $f(x)$. At $x = 4$, the term $\sqrt{4 - x}$ becomes zero, which is defined.

Thinking Process

A function $\dfrac{f(x)}{g(x)}$ is defined, if $g(x) \neq 0$.

Solution

Option (c): We have,

$ f(x)=\dfrac{4-x}{x-4} $

$f(x)$ is defined, if $x-4 \neq 0$ i.e., $x \neq 4$

$\therefore \quad$ Domain of $f=R- \lbrace 4 \rbrace $

Let $\quad f(x)=y$

$ \therefore \quad y=\dfrac{4-x}{x-4} \Rightarrow x y-4 y=4-x $

$\Rightarrow \quad x y+x=4+4 y $

$\Rightarrow \quad x(y+1)=4(1+y) $

$\therefore \quad x=\dfrac{4(1+y)}{y+1}$

$x$ assumes real values, if $y+1 \neq 0$ i.e., $y \neq-1$.

$\therefore \quad$ Range of $f=R- \lbrace -1 \rbrace $

• Option (a) Domain $=R$, Range $= \lbrace -1,1 \rbrace $: This is incorrect because the function $f(x)=\dfrac{4-x}{x-4}$ is not defined at $x=4$, so the domain cannot be all real numbers $R$. Additionally, the range is not limited to $ \lbrace -1, 1 \rbrace $; it includes all real numbers except $-1$.

• Option (b) Domain $=R- \lbrace 1 \rbrace $, Range $=R$: This is incorrect because the function $f(x)=\dfrac{4-x}{x-4}$ is not defined at $x=4$, not $x=1$. Therefore, the domain should exclude $4$, not $1$. Also, the range is not all real numbers $R$; it excludes $-1$.

• Option (d) Domain $=R- \lbrace -4 \rbrace $, Range $= \lbrace -1,1 \rbrace $: This is incorrect because the function $f(x)=\dfrac{4-x}{x-4}$ is not defined at $x=4$, not $x=-4$. Therefore, the domain should exclude $4$, not $-4$. Additionally, the range is not limited to $ \lbrace -1, 1 \rbrace $; it includes all real numbers except $-1$.

Solution

Option (d) We have $f(x)=\sqrt{x-1}$

The function $f$ is defined for

$x-1 \geq 0 \Rightarrow x \geq 1$

$\Rightarrow x \in [1 , \infty)$

$\therefore \quad $ Domain of $f = [1, \infty)$

For $x \geq 1$

$x-1 \geq 0 \Rightarrow \sqrt{x-1} \geq 0$

$f(x) \geq 0$

$\therefore $ Range of $f = [0, \infty)$

• Option (a) Domain $=(1, \infty)$, Range $=(0, \infty)$: This is incorrect because the domain should include 1, as $ f(x) = \sqrt{x-1} $ is defined for $ x = 1 $. Therefore, the domain should be $[1, \infty)$ instead of $(1, \infty)$.

• Option (b) Domain $=[1, \infty)$, Range $=(0, \infty)$: This is incorrect because the range $(0, \infty)$ incorrectly excludes $0,$ but $0$ is a value the function can take at $x=1$.

• Option (c) Domain $=(1, \infty)$, Range $=[0, \infty)$: This is incorrect because the domain should include 1, as $ f(x) = \sqrt{x-1} $ is defined for $ x = 1 $. Therefore, the domain should be $[1, \infty)$ instead of $(1, \infty)$.

Thinking Process

A function $\dfrac{f(x)}{g(x)}$ is defined, if $g(x) \neq 0$.

Solution

Option (a) We have, $f(x)= \dfrac{x^2+2x+1}{x^2-x-6}$

The function $f$ is defined if

$x^2-x-6 \neq 0$

$\Rightarrow \quad x^2-3x+2x-6 \neq 0$

$\Rightarrow \quad x(x-3)+2(x-3) \neq 0$

$\Rightarrow \quad x \neq 3, -2$

$\therefore \quad $ Domain of $f=R- \lbrace -2, 3 \rbrace $

• Option (b) $\mathbf{R}- \lbrace -3,2 \rbrace $: This is incorrect because this suggests that the domain excludes $x=-3 $ and $x=2$ but these value do not make the denominator zero.

• Option (c) $\mathbf{R}-[3,-2]$: This is incorrect because the interval notation implies a range of numbers, not specific points.

• Option (d) $\mathbf{R}-(3,-2)$: This is incorrect because this suggests excluding the open interval from $3 $ to $-2$. Similar to option C, this does not correctly represent the exclusion of specific points.

Solution

Option (b) We have, $f(x)= 2- \mid x-5 \mid$

Since, $\mid x-5 \mid $ is defined for all $x \in \mathbf{R}$.

$\therefore \quad f$ is defined for every $x \in \mathbf{R}$.

$\therefore \quad $ Domain of $f = \mathbf{R}$.

Now, we know that $\mid x-5 \mid \geq 0$

$\Rightarrow \quad - \mid x-5 \mid \leq 0$

$\Rightarrow \quad 2- \mid x-5 \mid \leq 2$

$\Rightarrow \quad f(x) \leq 2$

$\therefore \quad $ Range of $f= (- \infty , 2 ]$.

Option (a) Domain $=\mathbf{R}^{+}$, Range $=(-\infty, 1]$: This is incorrect because it states that the function is only defined for positive real numbers. Since $ f(x) = 2 - |x - 5| $ involves an absolute value, which is defined for all real $ x $, the domain should actually be all real numbers, $ \mathbf{R} $.

Option (c) Domain $=\mathbf{R}$, Range $=(-\infty, 2)$: This is incorrect because the range is given as $(- \infty, 2)$, suggesting that 2 is not included. However, the function reaches a maximum value of 2 at $ x = 5 $. Hence, 2 should be included in the range, making the correct range $(- \infty, 2]$.

Option (d) Domain $=\mathbf{R}^{+}$, Range $=(-\infty, 2]$: This is incorrect because it states that the function is only defined for positive real numbers. Since $ f(x) = 2 - |x - 5| $ involves an absolute value, which is defined for all real $ x $, the domain should actually be all real numbers, $ \mathbf{R} $.

Solution

Option (a) We have $f(x) = 3x^2-1$ and $g(x)=3+x$

$f(x) = g(x)$

$\Rightarrow \quad 3x^2-1 = 3+x$

$\Rightarrow \quad 3x^2-x-4 =0$

$\Rightarrow \quad 3x^2-4x+3x-4= 0$

$\Rightarrow \quad x(3x-4)+1(3x-4)=0$

$\Rightarrow \quad x=-1, \dfrac{4}{3}$

$\therefore \quad $ For $x \in \lbrace -1, \dfrac{4}{3} \rbrace , f(x)= g(x)$

Option (b) $\left[-1, \dfrac{4}{3}\right]$: This is incorrect because this suggests that $ f(x) $ and $ g(x) $ are equal for all $ x $ in the closed interval $\left[-1, \dfrac{4}{3}\right]$. However, the functions are only equal at the specific points $ x = -1 $ and $ x = \dfrac{4}{3} $, not throughout the interval.

Option (c) $\left(-1, \dfrac{4}{3}\right)$: This is incorrect because this suggests that the functions are equal for all $ x $ in the open interval $(-1, \dfrac{4}{3})$. However, the functions are not equal at any point in this interval except at the endpoints.

Option (d) $\left[-1, \dfrac{4}{3}\right)$: This is incorrect because this suggests that the functions are equal at $ x = -1 $ and for all $ x $ less than $\dfrac{4}{3}$. Again, the functions are only equal at the specific points $ x = -1 $ and $ x = \dfrac{4}{3} $.

Thinking Process

First find the domain of $f$ and domain of $g$. Then,

$ \text { domain of } f \cdot g=\text { domain of } f \cap \text { domain of } g \text {. } $

Solution

We have, $ f= \lbrace (0,1),(2,0),(3,-4),(4,2),(5,1) \rbrace $

And $g= \lbrace (1,0),(2,2),(3,-1),(4,4),(5,3) \rbrace $

$ \text { Domain of } f= \lbrace 0,2,3,4,5 \rbrace $

and Domain of $g= \lbrace 1,2,3,4,5 \rbrace $

$\therefore$ Domain of $(f \cdot g)=$ Domain of $ f \ \cap$ Domain of $g= \lbrace 2,3,4,5 \rbrace $

Solution

We have, $f= \lbrace (2,4),(5,6),(8,-1),(10,-3) \rbrace $

$\text { And } \quad g= \lbrace (2,5),(7,1),(8,4),(10,13),(11,5) \rbrace $

$\text { So, } f-g, f+g, f . g, \dfrac{f}{g} \text { are defined in the domain as} ( \text{domain of } f \cap \text { domain of } g ) $

$\text { i.e., } \lbrace 2,5,8,10 \rbrace \cap \lbrace 2,7,8,10,11 \rbrace = \lbrace 2,8,10 \rbrace $

$\text { (i) }(f-g)(2)=f(2)-g(2)=4-5=-1 $

$\quad (f-g)(8)=f(8)-g(8)=-1-4=-5 $

$\quad (f-g)(10)=f(10)-g(10)=-3-13=-16 $

$\therefore \quad f-g= \lbrace (2,-1),(8,-5),(10,-16) \rbrace $

(ii) $(f+g)(2)=f(2)+g(2)=4+5=9 $

$\quad (f+g)(8)=f(8)+g(8)=-1+4=3 $

$\quad (f+g)(10)=f(10)+g(10)=-3+13=10 $

$\therefore \quad f+g= \lbrace (2,9),(8,3),(10,10) \rbrace $

(iii) $(f \cdot g)(2)=f(2) \cdot g(2)=4 \times 5=20$

$ \begin{aligned} (f \cdot g)(8) & =f(8) \cdot g(8)=-1 \times 4=-4 \\ \\ (f \cdot g)(10) & =f(10) \cdot g(10)=-3 \times 13=-39 \\ \\ \therefore \quad f \cdot g & = \lbrace (2,20),(8,-4),(10,-39) \rbrace \end{aligned} $

(iv) $\dfrac{f}{g}(2)=\dfrac{f(2)}{g(2)}=\dfrac{4}{5}$

$\dfrac{f}{g}(8)=\dfrac{f(8)}{g(8)}=\dfrac{-1}{4}$

$\dfrac{f}{g}(10)=\dfrac{f(10)}{g(10)}=\dfrac{-3}{13}$

$\therefore \quad \dfrac{f}{g}= \bigg \lbrace \left(2, \dfrac{4}{5}\right), \left(8,-\dfrac{1}{4}\right), \left(10, \dfrac{-3}{13}\right) \bigg \rbrace $

Hence, the correct matches are (i) $\rightarrow$ (c), (ii) $\rightarrow$ (d), (iii) $\rightarrow$ (b), (iv) $\rightarrow$ (a).

Solution

False

We have, $\ R= \lbrace (x, y): y=x-5, x, y \in Z \rbrace $

If $\ x=5$, then $y=5-5=0$

Hence, $(5, 2)$ does not belong to $R$.

Solution

False

We have, $\ P= \lbrace 1,2 \rbrace $ and $n(P)=2$

$\therefore \quad n(P \times P \times P)=n(P) \times n(P) \times n(P)=2 \times 2 \times 2=8$

But given $P \times P \times P$ has 4 elements.

Thinking Process

First, we find $A \times B$ and $A \times C$, then we will find $(A \times B) \cup(A \times C)$.

Solution

True

We have, $ \ A= \lbrace 1,2,3 \rbrace , B= \lbrace 3,4 \rbrace $ and $C= \lbrace 4,5,6 \rbrace $

$A \times B= \lbrace (1,3),(1,4),(2,3),(2,4),(3,3),(3,4) \rbrace $

$A \times C= \lbrace (1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6) \rbrace $

$(A \times B) \cup(A \times C)= \lbrace (1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5)$, $(3,6) \rbrace $

Solution

False

We have, $ \ (x-2, y+5)=\left(-2, \dfrac{1}{3}\right)$

$\Rightarrow \quad x-2 =-2 \ \text{and} \ y+5=\dfrac{1}{3} $

$\Rightarrow \quad x=-2+2 \ \text{and} \ y=\dfrac{1}{3}-5 $

$\therefore \quad x =0 \ \text{and} \ y=\dfrac{-14}{3}$

Solution

True

We have, $\ A \times B= \lbrace (a, x),(a, y),(b, x),(b, y) \rbrace $

$A=$ Set of first element of ordered pairs in $A \times B= \lbrace a, b \rbrace $

$B=$ Set of second element of ordered pairs in $A \times B= \lbrace x, y \rbrace $